AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers 
represent the number of unit squares that have exactly 
edges on the boundary of respectively. Find the largest real number 
such that the inequality 
holds for each polygon constructed with these conditions.
such that, for all 
,
, the function 
is defined as, 
and for all 
, if the prime factorization of is 
then 
. We have an arithmetic sequence 
. If for a positive integer 
, the sequence 
is also an arithmetic sequence, show that the sequence 
has to be constant.
, 
are coprime positive integers. Find all 
such that:
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
vertices, each edge has one of the colors 
, 
, or 
. For each 
, if the vertices can be divided into 
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call 
to be a grandson of if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
, for which 
for all functions 
, who satisfy the equation 
for all non-negative real numbers and 
.
and 
be sequences of real numbers for which 
and
for all positive integers . Prove that is eventually increasing (that is, there exists a positive integer 
for which 
for all 
).
and 
be two concentric circles. Let 
and 
be any two equilateral triangles inscribed in and respectively. If and 
are any two points on and respectively, show that ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
such that the inequality 
holds for all positive integers 
.
be the assertion.
so 
, so 
, so 
, thus we have 
for all x.
, thus the only function is 
such that the inequality holds for all positive integers .
, you get 
and 
so . Now, we try to plug 
and 
respectively (not in once), and we get something like 
and 
. From the first equation it follows 
. Since we know that 
maps natural numbers to natural numbers, and we know that that equation works for all natural numbers , we can use our previous observation and plug 
, so we get something like 
so 
and 
must be 
. Ok, so now we look back again to previous equations and from it follows 
and from the 
so that means 
, but since 
, we get also for 
, combining these two things we get 
. Now from 
and we know 
so 
and hence is the only possibility. It's easy to check. Hope that helps
be the assertion of and to the given functional inequality. We will prove that only 
works. It is easy to check that this function holds.
.
gives us 
which forces 
since 
. Now, 
gives us 
, which forces .
for all 
gives us 
for all .
for all .![\[ f(x) + yf(x) \le f(x) + yf(f(x)) \le x(1 + f(y)) \Rightarrow \frac{f(x)}{x} \le \frac{f(y) + 1}{y + 1} \]](http://latex.artofproblemsolving.com/d/d/a/ddaa3f4c84e4f5a907c2e6c2ea3bc61ef1d02054.png)
Let 
, we then have for all . Therefore, we are done as we conclude 
for all .
such that the inequality holds for all positive integers .
denote the assertion of the given functional equation. 
gives us that 
, so 
. Next, 
gives us that ![\[f(y)-y\geq f(1)-1\geq0\]](http://latex.artofproblemsolving.com/e/4/9/e49df266eec87242fb895a6d68c4965e8cd90197.png)
so 
for all 
. Finally, we have ![\[xf(y)+x\geq yf(f(x))+f(x)\geq yf(x)+x\]](http://latex.artofproblemsolving.com/8/0/f/80fcb74e89905425d08efd1d72ac3af83143feb7.png)
so 
for all 
, interchanging and 
gives us that 
or 
for all where 
is a fixed constant. Plugging in gives us that ![\[(c-1)(cxy+x)\leq 0\]](http://latex.artofproblemsolving.com/e/9/6/e96d00b40a745fe57b28ec73509c629335b8512d.png)
and as 
, we must have 
, implying that 
for all 
.
is. and 
give 
Then gives 
Now it is not hard to get 
So 
denote the given assertion.
We know that because the functions takes in and out natural numbers.
we have 
Now, letting 
we have 
We also have 
using we have 
But we have 
therefore the only function that satisfies the given functional equation is 
.
From symmetry, 
. So 
, i. e. 
for some 
.
is the only working solution (by putting 
, for example), so the only solution is 
. 
combining above two relations we get
but we already have 
so f(x)=x is the only solution
such that 
Prove that
gives 
which gives 
Consequently, 
which then combined with 
Plugging 
whence we get 
.
yields 
implying that 
which again yields 
finishing the problem. 
.
,
for all 
combined with 
gives 
.
.
.
.
.
,
,
,
.
denote the given assertion.
.
.
.
by (*) 
.
.
.
for all 
.
.
.
.
.